$\lim _{h \rightarrow 0} \frac{ 2 \ln (1+h)-\ln (1+2 h)}{h^{2}}=$

Question

Professor PC Thomas Classes · Answer

$\lim _{h ightarrow 0} \frac{\ln (1+2 h)-2 \ln (1+h)}{h^{2}}$\ $=\lim _{h ightarrow 0} \frac{\ln (1+2 h)-\ln (1+h)^{2}}{h^{2}}$\ Considering $\ln (1+2 h)-\ln \left(1+2 h+h^{2}ight)$\ On expanding, we get\ $\lim _{h ightarrow 0} \frac{2 h-\left(2 h+h^{2}ight)}{h^{2}}=\lim _{h ightarrow 0}-\frac{h^{2}}{h^{2}}=-1$\